由网友(夏日夜.)分享简介：可能重复： What分配你从这个破碎的随机洗牌得到什么呢？ 这是从Skiena的算法设计手册。This is from Skiena's Algorithm Design Manual.假定myrand（A，B）产生之间的随机数a和b的包容性。Assume that myrand(a, b) generate...

可能重复：   What分配你从这个破碎的随机洗牌得到什么呢？

这是从Skiena的算法设计手册。

This is from Skiena's Algorithm Design Manual.

假定myrand（A，B）产生之间的随机数a和b的包容性。

Assume that myrand(a, b) generates a random number between a and b inclusive.

下面code产生均匀排列，随机

The following code generates permutations uniformly at random

for(int i = 0; i < n; i++)
    swap( a[i], a[myrand(i, n-1)]);

而以下则没有。

whereas the following doesn't.

for(int i = 0; i < n; i++)
    swap( a[i], a[myrand(0, n-1)]);

现在的问题是，为什么？

The question is, why?

推荐答案

在第一种情况下有恰好n！为执行该功能的可能路径（第一兰特有n的可能性，第二有n-1，...）。由于每个n个！可能的排列correspojnds到这些正好一个，它们被均匀地分布。

In the first case there are exactly n! possible paths for execution of the function (first rand has n possibilities, the second has n-1 ...). Since each of the n! possible permutations correspojnds to exactly one of these, they are uniformly distributed.

在第二种情况下，有n个分布^ n个可能的路径（n个可能chioces首次秒中，n ...）。这确实的没有的映射均匀的排列使分布是不均匀的。

In the second case, there are n^n possible paths of distribution (n possible chioces the first time, n the second...). This does not map uniformly to the permutations so the distribution is uneven.

要检查出来手动，生成所有可能性小N，就像3

To check it out "by hand", generate all possibilities for a small n, like 3

numbers chosen  generated permutation
(0,0,0)         (2,0,1)
...              ...